## College Algebra (6th Edition)

$\{\ (1,2),\ (1, -2),\ (-1,2),\ (-1, -2)\ \}$
1. Eliminating: eliminate the $(y^{2})$ terms: $\left\{\begin{array}{lll} 3x^{2}-2y^{2}=-5 & , & \\ 2x^{2}-y^{2}=-2 & , & /\times(-2), add \end{array}\right.$ $(3-4)x^{2}=-5+4$ $-x^{2}=-1$ 2. Solving: $-x^{2}=-1\qquad/\times(-1)$ $x^{2}=1$ $x=\pm 1$ 3. Back-substituting into $2x^{2}-y^{2}=-2$: $\left[\begin{array}{lll} x=1 & or & x=-1\\ 2(1)-y^{2}=-2 & & 2(1)-y^{2}=-2 \\ y^{2}=4 & & y^{2}=4\\ y=\pm 2 & & y=\pm 2 \end{array}\right]$ The solution set is $\{\ (1,2),\ (1, -2),\ (-1,2),\ (-1, -2)\ \}$